Some Important Discrete Probability Distributions

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Some Important Discrete

Probability Distributions

Learning Objectives

In this chapter, you learn:

The properties of a probability distribution

To calculate the expected value and variance of a probability

distribution To calculate the covariance and understand its use in finance

To calculate probabilities from binomial, hypergeometric,and Poisson distributions

How to use the binomial, hypergeometric, and Poissondistributions to solve business problems

DefinitionsRandom Variables

Random

Variables

Discrete

Random Variable

Continuous

Random VariableCh. 5 Ch. 6

Random

Variables

Discrete

Random Variable

Continuous

Random

Variables

Discrete

Random Variable

Continuous

Discrete Random Variables

Can only assume a countable number of values

Examples:

Roll a die twiceLet X be the number of times 4 occurs

(then X could be 0, 1, or 2 times)

Toss a coin 5 times.Let X be the number of heads

(then X = 0, 1, 2, 3, 4, or 5)

Probability Distribution For A Discrete

Random Variable

A probability distribution for a discrete random

variable is a mutually exclusive listing of all possible

numerical outcomes for that variable and a probability

of occurrence associated with each outcome.

Number of Classes Taken Probability

3 0.44 0.24

5 0.16

Example of a Discrete Random

Variable Probability Distribution

X Value Probability

0 1/4 = 0.25

1 2/4 = 0.50

2 1/4 = 0.25

Experiment: Toss 2 Coins. Let X = # heads.

4 possible outcomes

Probability Distribution

0 1 2 X

0.25 P r o b a b i l i t y

Discrete Random VariablesExpected Value (Measuring Center)

Expected Value (or mean) of a discrete

random variable (Weighted Average)

Example: Toss 2 coins,

X = # of heads,compute expected value of X:

E(X) = ((0)(0.25) + (1)(0.50) + (2)(0.25))

X P(X)

0 0.25

1 0.50

2 0.25

ii )X(PXE(X)

Discrete Random VariablesMeasuring Dispersion

Variance of a discrete random variable

Standard Deviation of a discrete random variable

where:

E(X) = Expected value of the discrete random variable X

Xi = the ith outcome of X

P(Xi) = Probability of the ith occurrence of X

2 )P(XE(X)][Xσ

)P(XE(X)][Xσσ

Discrete Random VariablesMeasuring Dispersion

Example: Toss 2 coins, X = # heads,compute standard deviation (recall E(X) = 1)

)P(XE(X)][Xσ i2

i 0.7070.50(0.25)1)(2(0.50)1)(1(0.25)1)(0σ 222

(continued)

Possible number of heads

= 0, 1, or 2

Covariance

The covariance measures the strength of the linear

relationship between two discrete random variables X

and Y.

A positive covariance indicates a positive relationship.

A negative covariance indicates a negative relationship.

The Covariance Formula

The covariance formula:

)YX(P)]Y(EY)][(X(EX[σ

1iiiiiXY

where: X = discrete random variable XXi = the ith outcome of XY = discrete random variable YYi = the ith outcome of YP(XiYi) = probability of occurrence of the

ith outcome of X and the ith outcome of Y

Investment Returns

The Mean

Economic Condition

Investment

Passive Fund X Aggressive Fund Y

0.2 Recession - $25 - $200

0.5 Stable Economy + $50 + $60

0.3 Expanding Economy + $100 + $350

Consider the return per $1000 for two types of

investments.

Investment Returns

The Mean

E(X) = μX = (-25)(.2) +(50)(.5) + (100)(.3) = 50

E(Y) = μY = (-200)(.2) +(60)(.5) + (350)(.3) = 95

Interpretation: Fund X is averaging a $50.00 return

and fund Y is averaging a $95.00 return per $1000invested.

Investment Returns

Standard Deviation

(.3)50)(100(.5)50)(50(.2)50)(-25σ222

71.193

)3(.)95350()5(.)9560()2(.)95200-(σ222

Interpretation: Even though fund Y has a higheraverage return, it is subject to much more variability

and the probability of loss is higher.

Investment Returns

Covariance

95)(.3)50)(350(100

95)(.5)50)(60(5095)(.2)200-50)((-25σXY

Interpretation: Since the covariance is large and

positive, there is a positive relationship between thetwo investment funds, meaning that they will likely

rise and fall together.

The Sum of Two Random Variables

Expected Value of the sum of two random variables:

Variance of the sum of two random variables:

Standard deviation of the sum of two random variables:

2YX σ2σσσY)Var(X

)Y(E)X(EY)E(X

2YXYX σσ

Portfolio Expected Return and Expected

Investment portfolios usually contain several different

funds (random variables)

The expected return and standard deviation of two

funds together can now be calculated.

Investment Objective: Maximize return (mean) whileminimizing risk (standard deviation).

Portfolio Expected Returnand Portfolio Risk

Portfolio expected return (weighted average return):

Portfolio risk (weighted variability)

Where w = proportion of portfolio value in asset X

(1 - w) = proportion of portfolio value in asset Y

)Y(E)w1()X(EwE(P)

2P w)σ-2w(1σ)w1(σwσ

Portfolio Example

Investment X: μX = 50 σX = 43.30

Investment Y: μY = 95 σY = 193.21

σXY = 8250

Suppose 40% of the portfolio is in Investment X and 60% is inInvestment Y:

The portfolio return and portfolio variability are between the values for

investments X and Y considered individually

77(95)(0.6)(50)0.4E(P)

133.30

)(8250)2(0.4)(0.6(193.71)(0.6)(43.30)(0.4)σ 2222P

Probability Distributions

Continuous

ProbabilityDistributions

Binomial

Hypergeometric

Poisson

Probability

Distributions

Discrete

ProbabilityDistributions

Normal

Uniform

Exponential

Ch. 5 Ch. 6

Binomial Probability Distribution

A fixed number of observations, n e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse

Each observation is categorized as to whether or not the

“event of interest” occurred e.g., head or tail in each toss of a coin; defective or not defective

light bulb

Since these two categories are mutually exclusive andcollectively exhaustive When the probability of the event of interest is represented as π,

then the probability of the event of interest not occurring is 1 - π

Constant probability for the event of interest occurring

(π) for each observation

Probability of getting a tail is the same each time we toss the

Binomial Probability Distribution

(continued)

Observations are independent The outcome of one observation does not affect the

outcome of the other

Two sampling methods deliver independence Infinite population without replacement

Finite population with replacement

Possible Applications for theBinomial Distribution

A manufacturing plant labels items as either

defective or acceptable

A firm bidding for contracts will either get a

contract or not

A marketing research firm receives survey

responses of “yes I will buy” or “no I will not”

New job applicants either accept the offer orreject it

The Binomial Distribution

Counting Techniques

Suppose the event of interest is obtaining heads on the toss

of a fair coin. You are to toss the coin three times. In how

many ways can you get two heads?

Possible ways: HHT, HTH, THH, so there are three ways you

can getting two heads.

This situation is fairly simple. We need to be able to count

the number of ways for more complicated situations.

C ti T h i

Counting Techniques

Rule of Combinations

The number of combinations of selecting X objects

out of n objects is

X)!(nX!

where: n! =(n)(n - 1)(n - 2) . . . (2)(1)

X! = (X)(X - 1)(X - 2) . . . (2)(1)

0! = 1 (by definition)

Counting Techniques

Rule of Combinations

How many possible 3 scoop combinations could you create

at an ice cream parlor if you have 31 flavors to select from?

The total choices is n = 31, and we select X = 3.

44952953128!123

28!293031

3)!(313!

31!C331

Binomial Distribution Formula

P(X) = probability of X events of interest in n trials, with the probability of an “event ofinterest” being π for each trial

X = number of “events of interest” in sample,

(X = 0, 1, 2, ..., n) n = sample size (number of trials

or observations)

π = probability of “event of interest”

X ! n Xπ (1-π)

X n X!

Example: Flip a coin fourtimes, let x = # heads:

π = 0.5

1 - π = (1 - 0.5) = 0.5

X = 0, 1, 2, 3, 4

Example:Calculating a Binomial Probability

What is the probability of one success in fiveobservations if the probability of an event ofinterest is .1?

X = 1, n = 5, and π = 0.1

0.32805

.9)(5)(0.1)(0

0.1)(1(0.1)1)!(51!

)(1X)!(nX!

n!1)P(X

Example

.01531

)(.8508)(45)(.0004

.02)(1(.02)2)!(102!

)(1X)!(nX!

n!2)P(X

Suppose the probability of purchasing a defective

computer is 0.02. What is the probability of

purchasing 2 defective computers in a group of 10?

X = 2, n = 10, and π = .02

Th Bi i l Di ib i

The shape of the binomialdistribution depends on thevalues of π and n

n = 5 π = 0.1

0 1 2 3 4 5 X

n = 5 π = 0.5

0 1 2 3 4 5 X

Here, n = 5 and π = .1

Here, n = 5 and π = .5

Th Bi i l Di t ib ti

Using Binomial Tables

n = 10

x … π=.20 π=.25 π=.30 π=.35 π=.40 π=.45 π=.50

… …

0.1074

0.2684

0.3020

0.2013

0.08810.0264

0.0055

0.0008

0.0001

0.0000

0.0563

0.1877

0.2816

0.2503

0.14600.0584

0.0162

0.0031

0.0004

0.0000

0.0282

0.1211

0.2335

0.2668

0.20010.1029

0.0368

0.0090

0.0014

0.0001

0.0000

0.0135

0.0725

0.1757

0.2522

0.23770.1536

0.0689

0.0212

0.0043

0.0005

0.0000

0.0060

0.0403

0.1209

0.2150

0.25080.2007

0.1115

0.0425

0.0106

0.0016

0.0001

0.0025

0.0207

0.0763

0.1665

0.23840.2340

0.1596

0.0746

0.0229

0.0042

0.0003

0.0010

0.0098

0.0439

0.1172

0.20510.2461

0.2051

0.1172

0.0439

0.0098

0.0010

… π=.80 π=.75 π=.70 π=.65 π=.60 π=.55 π=.50 x

Examples:

n = 10, π = .35, x = 3: P(x = 3|n =10, π = .35) = .2522

n = 10, π = .75, x = 2: P(x = 2|n =10, π = .75) = .0004

Binomial Distribution Characteristics

Variance and Standard Deviation

π nE(x)μ

)-(1nσ2

)-(1nσ π π Where n = sample size

π = probability of the event of interest for any trial

(1 – π) = probability of no event of interest for any trial

Th Bi i l Di t ib ti

Characteristics

n = 5 π = 0.1

0 1 2 3 4 5 X

n = 5 π = 0.5

0 1 2 3 4 5 X

0.5(5)(.1)nμ π

0.6708.1)(5)(.1)(1)-(1nσ

2.5(5)(.5)nμ π

.5)(5)(.5)(1)-(1nσ

Examples

U i E l F Th

Using Excel For The

Binomial Distribution

Th P i Di t ib ti

The Poisson Distribution

Definitions

You use the Poisson distribution when you areinterested in the number of times an event occurs in agiven area of opportunity.

An area of opportunity is a continuous unit orinterval of time, volume, or such area in which morethan one occurrence of an event can occur.

The number of scratches in a car’s paint

The number of mosquito bites on a person The number of computer crashes in a day

The Poisson Distribution

Apply the Poisson Distribution when:

You wish to count the number of times an event occurs in agiven area of opportunity

The probability that an event occurs in one area ofopportunity is the same for all areas of opportunity

The number of events that occur in one area of opportunityis independent of the number of events that occur in theother areas of opportunity

The probability that two or more events occur in an area ofopportunity approaches zero as the area of opportunitybecomes smaller

The average number of events per unit is (lambda)

Poisson Distribution Formula

where:

X = number of events in an area of opportunity

= expected number of events

e = base of the natural logarithm system (2.71828...)

Poisson Distribution Characteristics

Variance and Standard Deviation

λσ where = expected number of events

Using Poisson Tables

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

0.9048

0.0905

0.0045

0.00020.0000

0.0000

0.8187

0.1637

0.0164

0.00110.0001

0.0000

0.7408

0.2222

0.0333

0.00330.0003

0.0000

0.6703

0.2681

0.0536

0.00720.0007

0.0001

0.0000

0.6065

0.3033

0.0758

0.01260.0016

0.0002

0.0000

0.5488

0.3293

0.0988

0.01980.0030

0.0004

0.0000

0.4966

0.3476

0.1217

0.02840.0050

0.0007

0.0001

0.0000

0.4493

0.3595

0.1438

0.03830.0077

0.0012

0.0002

0.0000

0.4066

0.3659

0.1647

0.04940.0111

0.0020

0.0003

0.0000

Example: Find P(X = 2) if = 0.50

0.07582!

(0.50)e

e2)P(X

20.50Xλ

Using Excel For The

Poisson Distribution

Graph of Poisson Probabilities

0 1 2 3 4 5 6 7

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000P(X = 2) = 0.0758

Graphically:

= 0.50

Poisson Distribution Shape

The shape of the Poisson Distribution depends on

the parameter :

1 2 3 4 5 6 7 8 9 10 11 12

P ( x )

0 1 2 3 4 5 6 7

P ( x )

= 0.50 = 3.00

The Hypergeometric Distribution

The binomial distribution is applicable when selecting

from a finite population with replacement or from an

infinite population without replacement.

The hypergeometric distribution is applicable when

selecting from a finite population without replacement.

The Hypergeometric Distribution

“n” trials in a sample taken from a finite population of

size N

Sample taken without replacement

Outcomes of trials are dependent

Concerned with finding the probability of “X” items of

interest in the sample where there are “A” items of

interest in the population

Hypergeometric Distribution Formula

]C][C[P(X)

Xn ANX A

N = population size

A = number of items of interest in the populationN – A = number of events not of interest in the population

n = sample size

X = number of items of interest in the sample

n – X = number of events not of interest in the sample

Properties of theHypergeometric Distribution

The mean of the hypergeometric distribution is

The standard deviation is

Where is called the “Finite Population Correction Factor”

from sampling without replacement from a

finite population

nAE(x)μ

A)-nA(Nσ

U i th

Using theHypergeometric Distribution

■ Example: 3 different computers are checked out from 10 in

the department. 4 of the 10 computers have illegal software

loaded. What is the probability that 2 of the 3 selected

computers have illegal software loaded?

N = 10 n = 3

A = 4 X = 2

0.3120

(6)(6)

The probability that 2 of the 3 selected computers have illegal

software loaded is 0.30, or 30%.

Using Excel for the

Hypergeometric Distribution

Chapter Summary

Addressed the probability distribution of a discrete

random variable

Defined covariance and discussed its application in finance

Discussed the Binomial distribution

Discussed the Poisson distribution

Discussed the Hypergeometric distribution

Some Important Discrete Probability Distributions

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